Rate-Concentration Graphs

Spec Mapping — OCR H432 Module 5.1.1 — Rate–concentration graphs, covering the characteristic shapes of rate-vs-$[A]$[A] plots for zero, first and second order reactions (horizontal line, linear through origin, and upward parabola respectively), the extraction of the rate constant $k$k from the gradient of a first-order rate-$[A]$[A] plot, the linearisation of second-order data by plotting rate vs $[A]^2$[A]2, and the diagnostic role of these plots as the most cleanly discriminating graphical test of order (refer to the official OCR H432 specification document for exact wording).

Rate-concentration graphs are the cleanest visual diagnostic for order of reaction. Where concentration-time curves require careful half-life work to distinguish first from second order, a rate-vs-$[A]$[A] plot makes the distinction obvious by eye: a horizontal line says zero order; a straight line through the origin says first order; a parabolic upward curve says second order. This lesson develops the construction, interpretation, and use of these plots — including how to extract the rate constant $k$k directly from the gradient of a first-order plot, and how to linearise second-order data by plotting rate vs $[A]^2$[A]2. The rate-concentration approach complements the $[A]$[A]-$t$t tangent analysis of Lesson 3 and the half-life test of Lesson 5; combined, the three give triple-redundant evidence for the order of any reaction studied by continuous monitoring under PAG 9.

Key Equation: for each integer order, the rate-vs-$[A]$[A] relationship is $\text{Order 0:}\quad \text{rate} = k \quad \text{(horizontal line, y-intercept } = k\text{)}$Order 0:rate=k(horizontal line, y-intercept =k) $\text{Order 1:}\quad \text{rate} = k[A] \quad \text{(linear through origin, gradient } = k\text{)}$Order 1:rate=k[A](linear through origin, gradient =k) $\text{Order 2:}\quad \text{rate} = k[A]^2 \quad \text{(parabolic; linearise as rate vs }[A]^2\text{, gradient } = k\text{)}$Order 2:rate=k[A]2(parabolic; linearise as rate vs [A]2, gradient =k)

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What is a rate-concentration graph?

Whereas a concentration-time graph (Lesson 3) shows $[A]$[A] falling against time, a rate-concentration graph plots the instantaneous rate of reaction on the y-axis against the current concentration $[A]$[A] on the x-axis. The shape of this plot depends directly on the order:

• Zero order: horizontal straight line at height $k$k.
• First order: straight line passing through the origin with gradient $k$k.
• Second order: parabolic upward curve (rate $\propto [A]^2$∝[A]2).

Practical construction proceeds in two stages from a single PAG 9 continuous-monitoring experiment:

1. From the $[A]$[A] vs $t$t curve, draw tangents at several different times $t_i$ti​ and read off the gradients $-d[A]/dt$−d[A]/dt at each. The instantaneous rate at time $t_i$ti​ is the magnitude of that gradient.
2. Plot those rates against the corresponding $[A]_i$[A]i​ values from the curve. Each $(t_i, [A]_i)$(ti​,[A]i​) produces one $(\text{rate}_i, [A]_i)$(ratei​,[A]i​) point.

A single continuous-monitoring run thus yields enough data for a complete rate-$[A]$[A] plot.

Zero order — horizontal line

For zero order, rate = $k$k — rate does not depend on [A]. The plot is a horizontal straight line at height $k$k.

• The y-intercept gives $k$k directly (units: mol dm$^{-3}$−3 s$^{-1}$−1).
• The gradient is zero, confirming that [A] does not appear in the rate equation.
• As [A] falls during the reaction, the data points trace right-to-left along the same horizontal line.

First order — straight line through origin

For first order, rate = $k[A]$k[A]. The plot is a straight line passing through the origin with gradient $k$k.

• The line passes through the origin — when [A] = 0, rate = 0. (Strictly, in a real reaction [A] never reaches 0 exactly; we extrapolate.)
• The gradient equals $k$k (units: s$^{-1}$−1 for first order).
• This linear plot is the clearest evidence of first-order kinetics, alongside the constant-half-life signature (Lesson 5).

Worked example 1 — finding $k$k from a first-order plot

Q: A rate-[A] plot is straight, passing through the origin. At [A] = 0.50 mol dm$^{-3}$−3 the rate is 0.020 mol dm$^{-3}$−3 s$^{-1}$−1. Calculate $k$k.

A: gradient $k = \dfrac{\text{rate}}{[A]} = \dfrac{0.020}{0.50} = \mathbf{0.040\;\text{s}^{-1}}$k=[A]rate​=0.500.020​=0.040s−1.

Units check: (mol dm$^{-3}$−3 s$^{-1}$−1) / (mol dm$^{-3}$−3) = s$^{-1}$−1 — correct for first order.

Second order — upward parabolic curve

For second order, rate = $k[A]^2$k[A]2. The plot is a parabolic curve passing through the origin, starting shallow and becoming steeper as [A] increases.

• Not a straight line — distinguishes second from first order at a glance.
• To extract $k$k, replot rate vs $[A]^2$[A]2 instead — this is a straight line through the origin with gradient $k$k.
• Units of $k$k for second order: mol$^{-1}$−1 dm$^3$3 s$^{-1}$−1.

Worked example 2 — second-order linearisation

Q: From a continuous-monitoring run, the following rate-$[A]$[A] data are extracted:

[A] / mol dm$^{-3}$−3	rate / mol dm$^{-3}$−3 s$^{-1}$−1
0.10	$1.0 \times 10^{-3}$1.0×10−3
0.20	$4.0 \times 10^{-3}$4.0×10−3
0.30	$9.0 \times 10^{-3}$9.0×10−3
0.40	$1.6 \times 10^{-2}$1.6×10−2

(a) Confirm the order. (b) Find $k$k.

A: (a) Doubling [A] (0.10 → 0.20) takes rate from $1.0 \times 10^{-3}$1.0×10−3 to $4.0 \times 10^{-3}$4.0×10−3 — factor 4. Not first order ($\times 2$×2 would). Consistent with rate $\propto [A]^2$∝[A]2. Check: tripling [A] (0.10 → 0.30) takes rate $\times 9$×9 — confirms second order.

(b) Replot rate vs $[A]^2$[A]2:

$[A]^2$[A]2 / mol$^2$2 dm$^{-6}$−6	rate / mol dm$^{-3}$−3 s$^{-1}$−1
0.0100	$1.0 \times 10^{-3}$1.0×10−3
0.0400	$4.0 \times 10^{-3}$4.0×10−3
0.0900	$9.0 \times 10^{-3}$9.0×10−3
0.160	$1.6 \times 10^{-2}$1.6×10−2

Gradient = $(1.6 \times 10^{-2} - 1.0 \times 10^{-3})/(0.160 - 0.0100) = 0.015/0.150 = \mathbf{0.10\;\text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}}$(1.6×10−2−1.0×10−3)/(0.160−0.0100)=0.015/0.150=0.10mol−1dm3s−1.

Therefore rate = $0.10[A]^2$0.10[A]2 with $k = 0.10$k=0.10 mol$^{-1}$−1 dm$^3$3 s$^{-1}$−1.

Worked example 3 — zero-order detection

Q: A rate-$[A]$[A] table:

[A] / mol dm$^{-3}$−3	rate / mol dm$^{-3}$−3 s$^{-1}$−1
0.10	$2.0 \times 10^{-3}$2.0×10−3
0.20	$2.0 \times 10^{-3}$2.0×10−3
0.30	$2.0 \times 10^{-3}$2.0×10−3

Identify order and $k$k.

A: Rate is unchanged when [A] changes — flat horizontal line, zero order in A. $k = 2.0 \times 10^{-3}$k=2.0×10−3 mol dm$^{-3}$−3 s$^{-1}$−1 (the y-intercept).

Worked example 4 — first order with full check

Q: From a PAG 9 colorimetry experiment for $Br_2 + HCOOH \rightarrow 2HBr + CO_2$Br2​+HCOOH→2HBr+CO2​, the following data are extracted:

$[Br_2]$[Br2​] / mol dm$^{-3}$−3	rate / mol dm$^{-3}$−3 s$^{-1}$−1
0.0040	$5.6 \times 10^{-5}$5.6×10−5
0.0080	$1.12 \times 10^{-4}$1.12×10−4
0.0120	$1.68 \times 10^{-4}$1.68×10−4
0.0160	$2.24 \times 10^{-4}$2.24×10−4

(a) Order in $Br_2$Br2​? (b) $k$k?

A: (a) Doubling $[Br_2]$[Br2​] (0.0040 → 0.0080) doubles the rate ($5.6 \to 11.2 \times 10^{-5}$5.6→11.2×10−5). Tripling (0.0040 → 0.0120) triples it (5.6 → 16.8). First order in Br$_2$2​.

(b) Gradient of rate-vs-$[Br_2]$[Br2​] = $1.12 \times 10^{-4}/0.0080 = 0.014$1.12×10−4/0.0080=0.014 s$^{-1}$−1 (consistent across all four points). $k = \mathbf{1.4 \times 10^{-2}\;\text{s}^{-1}}$k=1.4×10−2s−1.

Summary table — three orders, three signatures

Order	rate-[A] plot	Gradient	y-intercept	Units of $k$k
0	Horizontal line	0	$k$k	mol dm$^{-3}$−3 s$^{-1}$−1
1	Straight line through origin	$k$k	0	s$^{-1}$−1
2	Parabolic upward curve	varies	0	mol$^{-1}$−1 dm$^3$3 s$^{-1}$−1
2 (linearised: rate vs $[A]^2$[A]2)	Straight line through origin	$k$k	0	mol$^{-1}$−1 dm$^3$3 s$^{-1}$−1

Why bother with rate-concentration graphs?

Concentration-time graphs (Lesson 3) reveal order through curve shape and half-life behaviour, but the first-vs-second order distinction can be subtle by eye. Rate-concentration graphs offer two major advantages:

1. Visual discrimination is unambiguous. A straight line through the origin (first order) is instantly distinguishable from a parabola (second order). Distinguishing curves on $[A]$[A]-$t$t requires multiple half-life measurements.
2. $k$k comes directly from the gradient. No half-life maths, no curve-fitting — just gradient = $k$k (for first-order) or linearise via $[A]^2$[A]2 for second-order.

The main disadvantage is that you need rate at multiple [A] values — which means drawing many tangents on the $[A]$[A]-$t$t curve. This propagates the tangent-drawing error into every point.

Worked example 5 — diagnostic decision tree

Q: A continuous-monitoring run gives this rate-$[A]$[A] table. Identify order and $k$k.

[A]	rate
0.10	$4.0 \times 10^{-3}$4.0×10−3
0.20	$1.6 \times 10^{-2}$1.6×10−2
0.40	$6.4 \times 10^{-2}$6.4×10−2

A: Doubling [A] (0.10 → 0.20) takes rate $\times 4$×4. Doubling again (0.20 → 0.40) takes rate $\times 4$×4. Both consistent with rate $\propto [A]^2$∝[A]2.

Therefore second order. Linearise: at [A] = 0.10, $[A]^2 = 0.01$[A]2=0.01, rate = $4.0 \times 10^{-3}$4.0×10−3, so $k = 4.0 \times 10^{-3}/0.01 = \mathbf{0.40\;\text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}}$k=4.0×10−3/0.01=0.40mol−1dm3s−1.

Cross-check at [A] = 0.40: $[A]^2 = 0.16$[A]2=0.16, rate = $6.4 \times 10^{-2}$6.4×10−2, $k = 6.4 \times 10^{-2}/0.16 = 0.40$k=6.4×10−2/0.16=0.40 — consistent.

Visualising the three shapes

graph TD
  A["Plot rate vs [A]"] --> B{Horizontal line?}
  B -->|Yes| C[Zero order; k = y-intercept]
  B -->|No| D{Straight line through origin?}
  D -->|Yes| E[First order; k = gradient]
  D -->|No| F["Plot rate vs [A]^2"]
  F --> G{Straight line through origin?}
  G -->|Yes| H[Second order; k = gradient]
  G -->|No| I[Try rate vs [A]^3 or log-log]

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Synoptic Links

Synoptic Links — Connects to:

• ocr-alevel-chemistry-quantitative-rates-equilibrium / rate-of-reaction-and-rate-equations (Lesson 1 — the rate equation that this graph linearises).
• ocr-alevel-chemistry-quantitative-rates-equilibrium / orders-of-reaction (Lesson 2 — the order definition this graph operationalises).
• ocr-alevel-chemistry-quantitative-rates-equilibrium / concentration-time-graphs (Lesson 3 — the source of the tangent-derived rate values plotted here).
• ocr-alevel-chemistry-quantitative-rates-equilibrium / rate-constant-k (Lesson 7 — units of $k$k derived dimensionally from each order's gradient).
• ocr-alevel-chemistry-quantitative-rates-equilibrium / the-arrhenius-equation (Lesson 9 — Arrhenius is another famous linearisation: $\ln k$lnk vs $1/T$1/T).
• aqa-alevel-mathematics-pure-2 / exponentials-and-logarithms (the maths of linearisation by logarithm is the same as for radioactive decay and bacterial growth).

Practical Activity Group anchor: PAG 9 (Continuous-monitoring rate) — the rate-concentration plot is constructed from tangent gradients on a PAG 9 $[A]$[A]-$t$t curve. Classic exemplars: $Br_2 + HCOOH$Br2​+HCOOH (colorimetry, first order in Br$_2$2​); Mg + HCl (gas-volume, first order in HCl when Mg in excess); decomposition of $NH_3$NH3​ on hot Pt or W (zero order at high [NH$_3$3​]). PAG 10 (Clock reactions) provides the alternative initial-rate approach — see Lesson 6.

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Specimen question modelled on the OCR H432 paper format

Question (9 marks): A student studied the reaction $A \rightarrow \text{products}$A→products by continuous monitoring. Drawing tangents to the resulting $[A]$[A] vs $t$t curve at several times produced the following rate-$[A]$[A] data:

[A] / mol dm$^{-3}$−3	rate / mol dm$^{-3}$−3 s$^{-1}$−1
0.500	$2.50 \times 10^{-2}$2.50×10−2
0.400	$1.60 \times 10^{-2}$1.60×10−2
0.300	$9.00 \times 10^{-3}$9.00×10−3
0.200	$4.00 \times 10^{-3}$4.00×10−3
0.100	$1.00 \times 10^{-3}$1.00×10−3

(a) Show that the reaction is second order in A. [3 marks]

(b) State what shape a plot of rate vs $[A]^2$[A]2 would have, and use it to calculate $k$k with units. [4 marks]

(c) Discuss why a rate-concentration plot is generally a better diagnostic of order than a concentration-time plot. [2 marks]

AO breakdown

Mark	AO	Awarded for
1	AO2	Doubling [A] from 0.10 to 0.20 takes rate × 4 (consistent with order 2)
2	AO2	Tripling [A] from 0.10 to 0.30 takes rate × 9 — second confirmation
3	AO3	Concludes: rate $\propto [A]^2$∝[A]2 at all five points, so order = 2
4	AO1	Plot of rate vs $[A]^2$[A]2 is straight line through origin
5	AO2	Gradient = $k$k = rate/$[A]^2$[A]2
6	AO2	Numerical value: $k = 0.100$k=0.100 (or 0.10) — consistent across at least two points
7	AO3	Units of $k$k: mol$^{-1}$−1 dm$^3$3 s$^{-1}$−1
8	AO3	Rate-conc plot gives straight lines that are visually distinct between orders
9	AO3	$k$k can be read directly from the gradient — avoids tangent-drawing error propagation